Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 779.
By properties of orthocenter, and cyclic quadrilateral, Angle DAC = Angle DHBThus, Triangle DAC ~ Triangle DHBNow, consider a rotation 90 degrees clockwise about point D, DB maps to a line segment on DC, DH maps to a line segment on DA. So, after the above rotation, triangle DHB will become the same position as triangle DAC, only with different size. Hence, DM will map to a line segment on DF. Which means, angle FDM = 90 degrees.
Triangle ADC is similar to triangle DHB And since F is the mid point of AC and M is the mid point of BH, so triangle ADF is similar to DHM and DFC is similar to DHMSo angle FDM is same as angle ADC , which is 90QED
http://img37.imageshack.us/img37/6047/problem779.pngDraw nine-points circle per attached sketchThis circle will pass through feet of altitudes, midpoint of each side and midpoint from orthocenter to each vertex.Since ∠ (MNF)=90 => MF is a diameterSo ∠ (MDF)=90
∆ADC is rt angled at D, and F is the midpoint of the hypotenuse AC.So AF = DF, ∠ADF = A.∆BDH is rt angled at D, and M is the midpoint of the hypotenuse BH.So ∠BDM = ∠DBM = 90° - A (note BH ⊥ AC)Follows ∠ADF + ∠BDM = 90°Hence ∠FDM = 90°